proof of Silverman-Toeplitz theorem

First, we shall show that the series $\sum_{n=0}^{\infty}a_{mn}z_{n}$ converges. Since the sequence $\{z_{n}\}$ converges, it must be bounded in absolute value — there must exist a constant $K>0$ such that $|z_{n}|\leq K$ for all $n$ . Hence, $|a_{mn}z_{n}|\leq K|a_{mn}|.$ Summing this gives

$\sum_{n=0}^{\infty}|a_{mn}z_{n}|\leq K\sum_{n=0}^{\infty}|a_{mn}|\leq KB.$

Hence, the series $\sum_{n=0}^{\infty}a_{mn}z_{n}$ is absolutely convergent which, in turn, implies that it converges.

Let $z$ denote the limit of the sequence $\{z_{n}\}$ as $n\to\infty$ . Then $|z|\leq K$ . We need to show that, for every $\epsilon>0$ , there exists an integer $M$ such that

$\left|\sum_{n=0}^{\infty}a_{mn}z_{n}-z\right|<\epsilon$

whenever $m>M$ .

Since the sequence $\{z_{n}\}$ converges, there must exist an integer $n_{1}$ such that $|z_{n}-z|<\frac{\epsilon}{4B}$ whenever $n>n_{1}$ .

By condition 3, there must exist constants $m_{o},m_{1},\ldots,m_{n_{1}}$ such that

$|a_{mn}|<{\epsilon\over 4(n_{1}+1)(K+1)}\quad\textrm{for}\quad 0\leq n\leq n_{% 1}\quad\textrm{and}\quad m>m_{n}.$

(1)

Choose $m^{\prime}=\max\{m_{0},m_{1},\ldots,m_{n_{1}}\}.$ Then

$\left|\sum_{n=0}^{n_{1}}a_{mn}z_{n}\right|\leq\sum_{n=0}^{n_{1}}|a_{mn}z_{n}|<% \sum_{n=0}^{n_{1}}{|z_{n}|\epsilon\over 4(n_{1}+1)(K+1)}<{\epsilon\over 4}$

(2)

when $m>m^{\prime}.$

By condition 2, there exists a constant $m^{\prime\prime}$ such that

$\left|\sum_{n=0}^{\infty}a_{mn}-1\right|<{\epsilon\over 6(|z|+1)}$

whenever $m>m^{\prime\prime}$ . By (1),

$\left|\sum_{n=0}^{n_{1}}a_{mn}\right|<{\epsilon\over 4(K+1)}\leq{\epsilon\over 4% (|z|+1)}$

when $m>m^{\prime}$ . Hence, if $m>m^{\prime}$ and $m>m^{\prime\prime}$ , we have

$\left|\sum_{n=n_{1}+1}^{\infty}a_{mn}-1\right|<{\epsilon\over 2(|z|+1)}.$

(3)

and

$\left|\sum_{n=n_{1}+1}^{\infty}a_{mn}(z_{n}-z)\right|\leq\sum_{n=n_{1}+1}^{% \infty}|a_{mn}|~{}|z_{n}-z|<{\epsilon\over 4B}\sum_{n=0}^{\infty}|a_{mn}|\leq{% \epsilon\over 4}.$

(4)

By the triangle inequality and (2), (3), (4) it follows that

$\left|\sum_{n=0}^{\infty}a_{mn}z_{n}-z\right|\leq\left|\sum_{n=0}^{n_{1}}a_{mn% }z_{n}\right|+\left|\sum_{n=n_{1}+1}^{\infty}a_{mn}(z_{n}-z)\right|+|z|\left|% \sum_{n=n_{1}+1}^{\infty}a_{mn}-1\right|$

$<\frac{\epsilon}{4}+\frac{\epsilon}{4}+\frac{\epsilon}{2}=\epsilon.$

Title	proof of Silverman-Toeplitz theorem
Canonical name	ProofOfSilvermanToeplitzTheorem
Date of creation	2013-03-22 14:51:35
Last modified on	2013-03-22 14:51:35
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	27
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 40B05